Difference between revisions of "2012 AMC 10B Problems/Problem 18"
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Therefore, <math>p</math> can be found by taking <math>\dfrac{1}{1+9.98}</math>, which is closest to <math>\dfrac{1}{1+10}</math> or <math>\dfrac{1}{11}</math>,or <math>\boxed{\textbf{(C)}}</math> | Therefore, <math>p</math> can be found by taking <math>\dfrac{1}{1+9.98}</math>, which is closest to <math>\dfrac{1}{1+10}</math> or <math>\dfrac{1}{11}</math>,or <math>\boxed{\textbf{(C)}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/igGLCogR-dk | ||
| + | |||
| + | ~savannahsolver | ||
== See Also == | == See Also == | ||
Revision as of 15:17, 30 January 2021
Contents
Problem
Suppose that one of every 500 people in a certain population has a particular disease, which displays no symptoms. A blood test is available for screening for this disease. For a person who has this disease, the test always turns out positive. For a person who does not have the disease, however, there is a 
false positive rate--in other words, for such people, 
of the time the test will turn out negative, but of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let 
be the probability that a person who is chosen at random from this population and gets a positive test result actually has the disease. Which of the following is closest to ?
Solution
This question can be solved by considering all the possibilities:
out of 
people will have the disease and will be tested positive for the disease.
Out of the remaining 
people, , or 
people, will be tested positive for the disease incorrectly.
Therefore, can be found by taking 
, which is closest to 
or 
,or 
Video Solution
~savannahsolver
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
